Tìm x
a |12,1.x+12,1.0,1|=12,1
/12,1.x+12,1.0,1/= 12,1
|12,1.x+12,1.0,1|=12,1
|12,1.x+12,1.0,1|=12,1
=> TH1: 12,1.x+12,1.0,1=12,1
(=) 12,1.(x+0,1) = 12,1
(=) x+0,1 = 12,1: 12,1
(=) x+0,1 = 1
(=) x = 1-0,1
(=) x = 0,9
TH2: 12,1.x+12,1.0,1= -12,1
(=) 12,1.(x+0,1) = -12,1
(=) x+0,1 = -1
(=) x = -1-0,1
(=) x = -1,1
Vậy, x = 0,9 hoặc x = -1,1
tìm x biết
a)/2,0.x-3,1/=6,3
b)12,1.x+12,1.0,1/=12,1
c)/0,2..x-3,1/+0,2,x+3,1/=0
x + x : 0,2 = 1,35
x * 1 + x * 5 = 1,35
x * ( 1 + 5 ) = 1,35
x * 6 = 1,35
x = 1,35 : 6
x = 0,225
hok tốt nha ^_^
Tìm x, biết
a) |0,2.x -3,1|= 6,3
b) |12,1.x +12,1.0,1| =12,1
c) |0,2.x - 3,1| +|0,2.x +3,1| =0
a) |0,2x - 3,1| = 6,3
\(0,2x-3,1=\pm6,3\)
Th1:
0,2x - 3,1 = 6,3
0,2x = 6,3 + 3,1
0,2x = 9,4
x = 9,4 : 0,2
x = 47
Th2:
0,2x - 3,1 = - 6,3
0,2x = - 6,3 + 3,1
0,2x = - 3,2
x = - 3,2 : 0,2
x = - 16
Vậy x = 47 hoặc x = - 16
b) |12,1x + 12,1 . 0,1| = 12,1
|12,1(x + 0,1)| = 12,1
\(12,1\left(x+0,1\right)=\pm12,1\)
Th1:
12,1(x + 0,1) = 12,1
x + 0,1 = 1
x = 1 - 0,1
x = 0,9
Th2:
12,1(x + 0,1) = - 12,1
x + 0,1 = - 1
x = - 1 - 0,1
x = - 1,1
Vậy x = 0,9 hoặc x = - 1,1
c) |0,2x - 3,1| + |0,2.x + 3,1| = 0
|0,2x - 3,1| + |0,2x + 3,1| \(\ge\) |0,2x - 3,1 + 0,2x + 3,1| = 0,4x
mà |0,2x - 3,1| + |0,2.x + 3,1| = 0
=> x = 0
tìm x biết
a)/0,2.x-3,1/=6,3
b)/12,1.x+12,1.0,1/=12,1
c)/0,2.x-3,1/+/0,2.x+3,1/=0
x + x : 0,2 = 1,35
x * 1 + x * 5 = 1,35
x * ( 1 + 5 ) = 1,35
x * 6 = 1,35
x = 1,35 : 6
x = 0,225
hok tốt nha ^_^
Tính nhanh
6,5+1,2+3,5-5,2+6,5-4,8
(-4,3.1,1+1,1.4,5).(-0,5.0,05+10,01)
(6,7+5,66-3,7+4,34).(-76,6.1,2+7,66.12)
Tìm x
|12,1.x+12,1.0,1|=12,1
|0,2.x-3,1|+|0,2+3,1|=0
bài 2)
a)|12,1x+12,1.0,1|=12,1
<=> |12,1(x+0,1)|=12,1
<=>\(\left[\begin{array}{nghiempt}12,1\left(x+0,1\right)=12,1\\12,1\left(x+0,1\right)=-12,1\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=0,9\\x=-1,1\end{array}\right.\)
b) |0,2x-3,1|+|0,2+3,1|=0
<=>|0,2x-3,1|+3,3=0
ta có |A|>=0
=> pt trên vô nghiệm
bài 1) a) 6,5+1,2+3,5-5,2+6,5-4,8
=(6,5+3,5+6,5)-(5,2-1,2+4,8)
=15,5-8,8=24,6
tt nha bạn
a) \(\left|12,1.x+12,1.0,1\right|=12,1\)
b) \(\left|0,2.x-3,1\right|+\left|0,2.x+3,1\right|=0\)
a/ \(\left|12,1x+12,1.0,1\right|=12,1\)
\(\Leftrightarrow\left|12,1.\left(x+0,1\right)\right|=12,1\)
\(\Leftrightarrow\left[{}\begin{matrix}12,1.\left(x+0,1\right)=12,1\\12,1.\left(x+0,1\right)=-12,1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+0,1=1\\x+0,1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0,9\\x=-1,1\end{matrix}\right.\)
Vậy ................
b/ \(\left|0,2x-3,1\right|+\left|0,2x+3,1\right|=0\)
Mà \(\left\{{}\begin{matrix}\left|0,2x-3,1\right|\ge0\\\left|0,2x+3,1\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|0,2x-3,1\right|=0\\\left|0,2x+3,1\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0,2x-3,1=0\\0,2x+3,1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}0,2x=3,1\\0,2x=-3,1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=15,5\\x=-15,5\end{matrix}\right.\)
Vậy ..
Tìm x biết:
a)\(\left|0,2x-3,1\right|=6,3\)
b)\(\left|12,1x+12,1.0,1\right|=12,1\)
c)\(\left|0,2x-3,1\right|+\left|0,2x+3,1\right|=0\)
mình biết làm đấy nhưng không biết ghi vào đây như thế nào!
/12,1 nhân x + 12,1 nhân 0,1/ =12,1